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<h3 class="heading"><span class="type">Paragraph</span></h3>
<p><dfn class="terminology">Examples</dfn>1. Solve the following system</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/Eq8_1.html">
\begin{equation*}
\begin{cases}
\frac{\partial u}{\partial t}=\frac{\partial^2 u}{\partial x^2},\quad x&gt;0, ~t&gt;0,\\
u(x, 0^+)=0,\quad x&gt;0\\
u(0, t)=\delta(t),\quad \lim \limits_{x \to \infty} u(x, t)=0,\quad t&gt;0
\end{cases}
\end{equation*}
</div>
<p class="continuation">(a) We begin by taking the Laplace transform, with respect to <span class="process-math">\(t\text{,}\)</span> of both sides</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/Eq8_1.html">
\begin{equation*}
{\mathcal L}[u_t(x, t)]=s {\mathcal L}[u(x, t)]={\mathcal L}[u_{xx}(x, t)].
\end{equation*}
</div>
<p class="continuation">Let <span class="process-math">\(L[u(x, t)]=U(x, s)\text{,}\)</span> then</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/Eq8_1.html">
\begin{equation}
s U=\frac{\mathrm{d}^2 U}{\mathrm{d} x^2} ~~\to~~ \frac{\mathrm{d}^2 U}{\mathrm{d} x^2} -s U=0.\tag{8.6.1}
\end{equation}
</div>
<p class="continuation">Notice that we have obtained an ODE for the unknown function <span class="process-math">\(U\text{.}\)</span>(b) The general solution of (<a href="" class="xref" data-knowl="./knowl/Eq8_1.html" title="Equation 8.6.1">(8.6.1)</a>) is</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/Eq8_1.html">
\begin{equation*}
U(x, s)=A(s) e^{\sqrt{s} x}+B(s) e^{-\sqrt{s}x}
\end{equation*}
</div>
<p class="continuation">Apply the boundary conditions, with <span class="process-math">\({\mathcal L}[f(t)]=F(s)\text{,}\)</span> we obtain</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/Eq8_1.html">
\begin{equation*}
U(0, s)={\mathcal L}[u(0, t)]={\mathcal L}(\delta(t))=1,
\end{equation*}
</div>
<p class="continuation">and</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/Eq8_1.html">
\begin{equation*}
\lim \limits_{x \to \infty} U(x, s)=\lim \limits_{x \to \infty} \int_0^{\infty} e^{-st} u(x, t) \mathrm{d} t=\int_0^{\infty} e^{-st} \lim \limits_{x \to \infty} u(x, t) \mathrm{d} t=0.
\end{equation*}
</div>
<p class="continuation">The boundary condition</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/Eq8_1.html">
\begin{equation*}
\lim \limits_{x \to \infty} U(x, s)=0 ~~\to~~ A(s)=0,
\end{equation*}
</div>
<p class="continuation">as for every fixed <span class="process-math">\(s&gt;0\text{,}\)</span> <span class="process-math">\(e^{\sqrt{s} x}\)</span> increases as <span class="process-math">\(x \to \infty\text{.}\)</span> Hence</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/Eq8_1.html">
\begin{equation*}
U(0, s)=B(s)=1.
\end{equation*}
</div>
<p class="continuation">Therefore,</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/Eq8_1.html">
\begin{equation*}
U(x, s)=e^{-\sqrt{s}x}.
\end{equation*}
</div>
<p class="continuation">(c) From the table of Laplace transform, we obtain the inverse Laplace transform as</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/Eq8_1.html">
\begin{equation*}
{\mathcal L}^{-1}(e^{-\sqrt{s} x})=\frac{x}{2 \sqrt{ \pi t^3}} e^{-\frac{x^2}{4 t}}.
\end{equation*}
</div>
<p class="continuation">Hence</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/Eq8_1.html">
\begin{equation*}
u(x, t)=\frac{x}{2 \sqrt{\pi t^3}} e^{-\frac{x^2}{4 t}}.
\end{equation*}
</div>
<span class="incontext"><a href="sec8_6.html#p-503" class="internal">in-context</a></span>
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